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Leto [7]
3 years ago
5

WHAT THE HECK IS THAT SHAPE PLEASE HELP IM SLOW

Mathematics
2 answers:
OLga [1]3 years ago
7 0

Answer:

trapezoid

Step-by-step explanation

there are two lines across from each toher that are parallel, but the other two lines aren't , so this has to be a trapezoid. there are also no right angles

SCORPION-xisa [38]3 years ago
5 0

Answer: It looks like a Trapezoid

Step-by-step explanation:

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3 years ago
Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
Adrianne invested $2000 in an account at a 3.5% interest rate compounded annually. She made no deposits or withdrawals on the ac
Musya8 [376]

Answer:

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Step-by-step explanation:

This is our equation that I could find within this question.

2000(1+.035)^4

==> 2000(1.035)^4

==> 2000 * 1.1475

==> 2,295

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3 years ago
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never [62]

Answer:

Exact Form:

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Decimal Form:

1.3

Mixed Number Form:

1 1/3

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